The correct order of reduction potentials of the following pairs is
A. $$Cl_2/Cl^-$$
B. $$I_2/I^-$$
C. $$Ag^+/Ag$$
D. $$Na^+/Na$$
E. $$Li^+/Li$$
Choose the correct answer from the options given below.

The standard reduction potential, $$E^\circ$$, indicates the tendency of a species to gain electrons and undergo reduction. A more positive value of $$E^\circ$$ corresponds to a greater tendency for reduction.

The given couples are

• A: $$Cl_2/Cl^-$$
• B: $$I_2/I^-$$
• C: $$Ag^+/Ag$$
• D: $$Na^+/Na$$
• E: $$Li^+/Li$$

The standard reduction potentials of these couples are approximately

$$E^\circ(Cl_2/Cl^-)=+1.36\text{ V}$$

$$E^\circ(Ag^+/Ag)=+0.80\text{ V}$$

$$E^\circ(I_2/I^-)=+0.54\text{ V}$$

$$E^\circ(Na^+/Na)=-2.71\text{ V}$$

$$E^\circ(Li^+/Li)=-3.04\text{ V}$$

Among the halogens, chlorine has a higher reduction potential than iodine because chlorine is a stronger oxidizing agent.

$$Cl_2/Cl^- > I_2/I^-$$

Silver is a noble metal and has a positive reduction potential, which lies between those of chlorine and iodine.

$$Cl_2/Cl^- > Ag^+/Ag > I_2/I^-$$

Alkali metals possess highly negative reduction potentials because they readily lose electrons. Lithium has the most negative reduction potential due to the exceptionally high hydration enthalpy of $$Li^+$$.

$$Na^+/Na > Li^+/Li$$

Combining all the values,

$$Cl_2/Cl^- > Ag^+/Ag > I_2/I^- > Na^+/Na > Li^+/Li$$

Hence, the correct order is

$$\boxed{A>C>B>D>E}$$

Therefore, the correct answer is Option D.